I have the following probability function:

$$\text{Prob} = \frac{1}{1 + e^{-z}}$$


$$z = B_0 + B_1X_1 + \dots + B_nX_n.$$

My model looks like

$$\Pr(Y=1) = \frac{1}{1 + \exp\left(-[-3.92 + 0.014\times(\text{bid})]\right)}$$

This is visualized via a probability curve which looks like the one below.

enter image description here

I am considering adding a couple variables to my original regression equation. Let's say I add gender (categorical: F and M) and age (categorical: < 25 and > 26) into the model, I end up with:

$$\Pr(Y=1) = \frac{1}{1 + \exp\left(-[-3.92 + 0.014\times(\text{bid}) + 0.25\times(\text{gender}) + 0.15\times(\text{age})]\right)}$$

In R I can generate a similar probability curve which will tell me the probability of Y=1 when accounting for all three predictors. Where I'm lost is I want to find the probabilities for every possible permutation of these variations.

So when bid = 1, gender = M, and age is >= 26, what is the probability that Y = 1? Similarly, when bid = 2, gender = F, and age is >= 26, what is the probability that Y = 1?

I want to generate a probability curve which will allow me to visualize this.

Can anyone help? I may be completely misunderstanding what kind of information one can glean from a logit model, but please tell me if I am also misunderstanding the theory.

  • Do you want the code to do this in R, or just to understand the issue conceptually? – gung Jul 3 '12 at 21:50
  • If I had to choose, I'd say the issue conceptually. I think the R code I can mange to figure out on my own. – ATMathew Jul 3 '12 at 21:56
  • 2
    If you are comfortable solving the same problem with ordinary (least squares) regression, then why not just express the response as log odds (which is just $B_0 + B_1X_1 + \dots + B_nX_n$) and use the techniques you know? – whuber Jul 3 '12 at 22:16
  • 1
    Sure, take a look at @FrankHarrell rms package (extensive documentation can be found on the RMS website). Start with the Predict() and plot.Predict() functions to get a feel of what can be done (this includes plotting $\Pr(Y=1|x_2,\dots,x_p)$ as a function of $x_1$, with $x_2,\dots,x_p$ set at default values, or fixed values of your choice). – chl Jul 4 '12 at 20:09
up vote 21 down vote accepted

Fortunately for you, you have only one continuous covariate. Thus, you can just make four (i.e., 2 SEX x 2 AGE) plots, each with the relation between BID and $p(Y=1)$. Alternatively, you could make one plot with four different lines on it (you could use different line styles, weights, or colors to distinguish them). You can get these predicted lines by solving the regression equation at each of the four combinations for a range of BID values.

A more complicated situation is where you have more than one continuous covariate. In a case like this, often there is a particular covariate that is 'primary' in some sense. That covariate can be used for the X axis. Then you solve for several pre-specified values of the other covariates, typically the mean and +/- 1SD. Other options include various types of 3D plots, coplots, or interactive plots.

My answer to a different question here has information on a range of plots for exploring data in more than 2 dimensions. Your case is essentially analogous, except that you are interested in presenting the model's predicted values, rather than the raw values.


I have written some simple example code in R to make these plots. Let me note a few things: Because the 'action' takes place early, I only ran BID through 700 (but feel free to extend it to 2000). In this example, I am using the function you specify and taking the first category (i.e., female and young) as the reference category (which is the default in R). As @whuber notes in his comment, LR models are linear in log odds, thus you can use the first block of predicted values and plot as you might with OLS regression if you choose. The logit is the link function, which allows you to connect the model to probabilities; the second block converts log odds into probabilities via the inverse of the logit function, that is, by exponentiating (turning into odds) and then dividing the odds by 1+odds. (I discuss the nature of link functions and this type of model here, if you want more info.)

BID = seq(from=0, to=700, by=10)

logOdds.F.young = -3.92 + .014*BID
logOdds.M.young = -3.92 + .014*BID + .25*1
logOdds.F.old   = -3.92 + .014*BID         + .15*1
logOdds.M.old   = -3.92 + .014*BID + .25*1 + .15*1

pY.F.young = exp(logOdds.F.young)/(1+ exp(logOdds.F.young))
pY.M.young = exp(logOdds.M.young)/(1+ exp(logOdds.M.young))
pY.F.old   = exp(logOdds.F.old)  /(1+ exp(logOdds.F.old))
pY.M.old   = exp(logOdds.M.old)  /(1+ exp(logOdds.M.old))

  plot(x=BID, y=pY.F.young, type="l", col="blue", lwd=2, 
       ylab="Pr(Y=1)", main="predicted probabilities for young women")
  plot(x=BID, y=pY.M.young, type="l", col="blue", lwd=2, 
       ylab="Pr(Y=1)", main="predicted probabilities for young men")
  plot(x=BID, y=pY.F.old, type="l", col="blue", lwd=2, 
       ylab="Pr(Y=1)", main="predicted probabilities for old women")
  plot(x=BID, y=pY.M.old, type="l", col="blue", lwd=2, 
       ylab="Pr(Y=1)", main="predicted probabilities for old men")

Which produces the following plot:
enter image description here
These functions are sufficiently similar that the four-parallel plot approach I outlined initially is not very distinctive. The following code implements my 'alternative' approach:

  plot(x=BID, y=pY.F.young, type="l", col="red", lwd=1, 
       ylab="Pr(Y=1)", main="predicted probabilities")
  lines(x=BID, y=pY.M.young, col="blue", lwd=1)
  lines(x=BID, y=pY.F.old,   col="red",  lwd=2, lty="dotted")
  lines(x=BID, y=pY.M.old,   col="blue", lwd=2, lty="dotted")
  legend("bottomright", legend=c("young women", "young men", 
         "old women", "old men"), lty=c("solid", "solid", "dotted",
         "dotted"), lwd=c(1,1,2,2), col=c("red", "blue", "red", "blue"))

producing in turn, this plot:
enter image description here

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Not the answer you're looking for? Browse other questions tagged or ask your own question.